Mtan

[Ryan Rigdon]

y = 1/(x+6)

find the equation of the tangent line at (-9,-1/3)

1st find the slope.

my work

Mtan = lim (h to 0) (f(-9+h) -f(9))/h = lim (h to 0) ((1)/(-3+h)+1/3)/h

lim (h to 0) ((h)/((-3+h)(3)))/h = lim (h to 0) 1/((-3+h)(3)) = -1/9


I think this is correct, however when I use (3+h)(3) instead i get 1/9. Is my work accurate?

 

[Ryan Rigdon]

I also used (3-h)(3) as the denominator (did this because i had (-3+h)) on the top part of the lim and got the same result of -1/9.

with this said i think that my answer of Mtan = -1/9 is correct. Just want to make sure before i move on to the next step.

 

[Deleted member 4993]

y = 1/(x+6)

find the equation of the tangent line at (-9,-1/3)

1st find the slope.

my work

Mtan = lim (h to 0) (f(-9+h) -f(9))/h = lim (h to 0) ((1)/(-3+h)+1/3)/h

lim (h to 0) ((h)/((-3+h)(3)))/h = lim (h to 0) 1/((-3+h)(3)) = -1/9 <<< You can check this by plotting the curve and observing the slope of the tangent line at x = -9

Spoiler: :35rylqz6

Slope = - 1/9 is correct[/spoiler:35rylqz6]
I think this is correct, however when I use (3+h)(3) instead i get 1/9. Is my work accurate?

 

[Ryan Rigdon]

thanx again for the help. one thing i have learned in mathematics is that especially in calculus that there is more than one way to find the correct answer.

 

[galactus]

I see you worked out this derivative by first principles. Are you up to the product or quotient rules yet?.

Product rule:

\(\displaystyle \frac{d}{dx}\left[\frac{1}{x+6}\right]=\frac{d}{dx}[(x+6)^{-1}]=-(x+6)^{-2}=\frac{-1}{(x+6)^{2}}\)

Quotient rule:

\(\displaystyle \frac{d}{dx}\left[\frac{1}{x+6}\right]=\frac{(x+6)(0)-1(1)}{(x+6)^{2}}=\frac{-1}{(x+6)^{2}}\)

See, both yield the same result.

Now, just sub in x=-9 and get the slope of the tangent line at that point.

Something to remember is that one can most often use the product rule over the quotient rule because it is easier to apply in most cases.

In this one I just rewrite \(\displaystyle \frac{1}{x+6}\) as \(\displaystyle (x+6)^{-1}\)

Just thought I would throw this out there.